Sickle-cell anemia is an inherited disease in which red blood cells are misshape

Question

Sickle-cell anemia is an inherited disease in which red blood cells are misshapen and sticky. Sickle cells tend to form clumps in blood vessels, inhibiting the flow of blood. Humans have two genes for sickle-cell anemia, either of which may be S for normal cells or s for sickle cells. A person with two copies of the gene s will have sickle-cell anemia. A person with one s gene and one S gene will not have the disease, but will be a carrier, which means that the s gene may be transmitted to a person's offspring. If two carriers have a child, the probability is .25 that the child will have the disease and 0.5 that the child will be a carrier. Outcomes among children are independent.
a) A mother and father who are both carriers have two children. What is the probability that neither child has the disease?
b) What is the probability that both children are carriers?
c) If neither child has the disease, what is the probability that both are carriers?
d) A woman who is the child of two carriers has a child by a man who is a carrier. What is the probability that the child has the disease?

Explanation / Answer

a) If you make a Punnet square for 2 carrier parents (Aa x Aa), you get AA, Aa, aA, aa. there is only one child with the disease so that is why it's 1/4=.25 probability. So the probability of two child being either homozygous dominant (AA) or heterozygous (Aa or aA) is (3/4)*(3/4) = 9/16. They have to be multiplied because it's not an 'either-or' senario. (look up 'the sum rule' and 'the product rule' on probability, if you dont understand this part) b) carriers are heterozygous. So there are two possible genotypes (Aa, aA) so it's 1/2 chance for each child. (1/2)*(1/2) = 1/4. c) Since you KNOW that they are either homozygous dominant or heterozygous. So instead of 4 possibilities (AA, Aa, aA, aa), there are only 3 (AA, Aa, aA). But there are still 2 possible genotypes for heterozygous so (2/3)*(2/3) = 4/9. d) For this one, you have to calculate the probability of the mother and the child. The woman is a child of 2 carriers. The problem doesnt say that she has the disease so it means that she is heterozygous. The probability of the woman being heterozygous is (2/3)*(2/3) = 4/9. You know that the man is a carrier. Now, it's a cross between heterozygous parents. Meaning, (1/2)*(1/2) = 1/4. All you need to do now is multiply them. (4/9)*(1/4) = 1/9.

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